Question 755056
If there is a {{{ sqrt(-1) }}} in the solutions, then
the solutions are not all real.
(a)
{{{ x^2 - 9 = 0 }}}
Add {{{9}}} to both sides
{{{ x^2 = 9 }}}
{{{ x = 3 }}}
and
{{{ x = -3 }}}
Both are real solutions
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(b)
{{{ x^2 + 7 = 1 }}}
Subtract {{{ 7 }}} from both sides
{{{ x^2 = -6 }}}
{{{ x = sqrt(-1)*sqrt(6) }}}
{{{ x = sqrt(6)*i }}}
and also,
{{{ x = -( sqrt(-1)*sqrt(6) ) }}}
{{{ x = -sqrt(6)*i }}}
There are no real solutions
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(c)
{{{ 3x - 7 = 14 }}}
{{{ 3x = 21 }}}
{{{ x = 7 }}}
There is 1 real solution